2025 AMC 10B Problem 14

Attempt Problem 14 of the 2025 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 10B solutions, or check the answer key.

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14.

Nine athletes, no two of whom are the same height, try out for the basketball team. One at a time, they draw a wristband at random, without replacement, from a bag containing 33 blue bands, 33 red bands, and 33 green bands. They are divided into a blue group, a red group, and a green group. The tallest member of each group is named the group captain. What is the probability that the group captains are the three tallest athletes?

29\dfrac{2}{9}

27\dfrac{2}{7}

928\dfrac{9}{28}

13\dfrac{1}{3}

38\dfrac{3}{8}

Answer: C
Concepts:basic probabilityconditional probability
Difficulty rating: 1500
Solution:

The captains are the three tallest exactly when those three land in three different groups, since each is then the tallest of its own group. Drop them into the 99 slots one at a time (33 per group). The second tallest misses the first's group with probability 68,\tfrac{6}{8}, and the third misses both with probability 37.\tfrac{3}{7}. So the probability is 6837=928.\tfrac{6}{8} \cdot \tfrac{3}{7} = \tfrac{9}{28}. Therefore, the answer is C.

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